An LED (light-emitting diode) requires a voltage of 2.0 V across it and a current of 20 mA to operate. It is connected to a 9.0 V battery with a protective ...
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An LED (light-emitting diode) requires a voltage of 2.0 V across it and a current of 20 mA to operate. It is connected to a 9.0 V battery with a protective resistor R in series.
(a) Calculate the voltage across the protective resistor R. [1]
(b) Calculate the resistance of R. [2]
(c) Explain why the protective resistor is necessary. [1]
(d) State one advantage of using an LED instead of a filament lamp. [1]
Answer Details
(a) The LED needs 2.0 V and the battery supplies 9.0 V. In a series circuit, voltages add up to the supply voltage, so the voltage across the protective resistor R is:
\( V_R = 9.0 - 2.0 = \mathbf{7.0 \text{ V}} \) [1]
(b) The current through the LED is 20 mA, which must be converted to amps:
\( I = 20 \text{ mA} = 0.020 \text{ A} \)
Since R and the LED are in series, the same current flows through both. Using \( R = \frac{V}{I} \):
\( R = \frac{7.0}{0.020} \) [1]
\( R = \mathbf{350 \;\Omega} \) [1]
(c) The protective resistor is necessary because, without it, the full 9.0 V battery voltage would be applied directly across the LED. This would drive too much current through the LED, exceeding its maximum rating and destroying it. [1]
The resistor limits the current to the safe operating value (20 mA) by "dropping" the excess voltage (7.0 V out of 9.0 V).
(d) One advantage of using an LED instead of a filament lamp: LEDs are more energy-efficient (they convert a greater proportion of electrical energy into light rather than wasting it as heat). [1]
Other valid advantages include: LEDs last much longer, produce less heat, switch on instantly, and are physically smaller.
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